Accumulation points of point spectrum of Schrödinger operator in one dimension

Question

Consider a Schrödinger operator $H=-\partial_x^2+V(x)$, with $x\in\mathbb R$, $V(x)$ tending monotonically to $V_\pm$ as $x\to\pm\infty$, and $\min V(x)<V\pm$. Intuitively, the only accumulation point of the point spectrum of $H$ must belong to the continuous spectrum, and thus coincide with $\min\{V_\pm\}$. How can this be rigorously proved, and is this fact a consequence of some general principle about the spectrum of a self-adjoins operator?

Answer 1

This follows indeed from a number of standard results, though a written out proof with all details included would be lengthy.

(1) The essential spectrum (this, and not continuous spectrum is the proper term in mathematical usage when referring to the set of accumulation points of the spectrum [and eigenvalues of infinite multiplicity, which we can't have here) of the whole line problem is the union of those of the two half line problems $H_{\pm}$, and $\sigma_{ess}(H_{\pm})=[V_{\pm},\infty)$. Thus $\sigma_{ess}(H)=[\operatorname{min} (V_-,V_+),\infty)$.

(2) If $V$ is monotone (or, more generally, of bounded variation), the essential spectrum is purely absolutely continuous. This is most naturally done in a half line setting first, but we obtain the same conclusion for the whole line problem because we would need solutions with asymptotic decay for singular spectrum, and the proof shows that instead the solutions obey WKB asymptotics.

As a consequence, the spectrum is purely absolutely continuous on $[\operatorname{min} (V_-,V_+),\infty)$, and eigenvalues can only accumulate at $\operatorname{min} (V_-,V_+)$.